Question about "Coding the universe" The following is a result which I know as a weak form of Jensen's coding lemma$^*$  (first published in the book "Coding the universe"; also see http://www.jstor.org/stable/2273986):

For any class of ordinals $A\subseteq ON$ with $V=L[A]$, there is a class forcing $\mathcal{P}$ such that $\Vdash_\mathcal{P}V=L[a], a\subseteq\omega$.

(The coding theorem says more than this, but I doubt that the claim above has a substantially easier proof.)
My question is:


*

*Can the class forcing $\mathcal{P}$ can assumed to be a set forcing, in the case when $A$ is a set?


I suspect this is true; however, the proof of the coding lemma is too difficult for me to untangle easily, so I'm not sure this is true.

$^*$ I've heard this called "Jensen's Coding Theorem," but the book is by Jensen and Beller & Welch; is this indeed due to Jensen, or all three of them?
 A: According to how I understand your question, the answer to your question is no. If $V$ is not of the form $L[x]$, where $x$ is a set, then it is impossible to have a set-forcing extension $V[G]$ of the form $L[a]$, where $a$ is a set. If $V[G]=L[a]$, where $a$ is a set and $G\subset P\in V$ is $V$-generic, then fix a name $\dot a$ for $a$, and consider $L[P][\dot a]$ (one should use well-ordered versions of these names to have a ZFC model). So $G$ is also $L[P][\dot a]$-generic, and $L[P][\dot a][G]$ has $a$ and hence includes $L[a]$ and is contained in $L[a]$, and so $L[P][\dot a][G]=L[a]=V[G]$. In particular, this means $$L[P][\dot a]\subset V\subset V[G]=L[P][\dot a][G]=L[a],$$ and so by the intermediate model theorem, $V$ must be a set-forcing extension of $L[p][\dot a]$, and so $V=L[P][\dot a][H]$ for some $H$ generic for a subforcing of $P$. In particular, this means that $V$ has the form $L[x]$ for the set $x=(P,\dot a,H)$, contrary to assumption. 
It is easy to make models $V$ not of the form $L[x]$, simply by performing any of the usual class forcing, such as the usual Easton forcing or the canonical forcing of the GCH. 
But I think that perhaps you meant to say that $V=L[A]$ as the hypothesis about $A$. In this case, if $A$ is a set, then of course we can make $V[G]=L[a]$ for $a\subset\omega$, simply by collapsing $\sup A$ to $\omega$ and then coding both $A$ and the collapsing function with a real. So under that interpretation of the question, the answer would be positive.
